I found this problem on this site: problem 3
I was thinking maybe I can prove this by contradiction. Assuming all paths $P_1,\cdots$ share at least one vertex which then would contradict the hypothesis that there is no vertex in $V(G)-s,t$ that is in more than 2 paths $Q_1,\cdots$. is there any merit to this? how can one prove it? thank you.
Assuming your are allowed to use Menger's theorem, we can go the way over connectivity.
Proof.
Take $k-1$ vertices except $s$ or $t$. Removing them can at most destroy $2k-2$ of the paths $Q_i$. Hence $s$ and $t$ are still connected after the removal.
This means that the smallest seperator of $s$ and $t$ has $k$ vertices, which implies the existence of the paths $P_1,...,P_k$ by Menger's theorem.